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Timeline for Partitioning a convex $n$-polygon

Current License: CC BY-SA 4.0

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May 2, 2021 at 11:15 history edited TheVal CC BY-SA 4.0
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May 2, 2021 at 10:14 vote accept TheVal
S Apr 30, 2021 at 20:53 history suggested Jukka Kohonen CC BY-SA 4.0
attempted to clarify the question, as I understand from the comments. In particular, parts are NOT disjoint but their interiors are.
Apr 30, 2021 at 13:40 review Suggested edits
S Apr 30, 2021 at 20:53
Apr 30, 2021 at 13:20 answer added Joseph O'Rourke timeline score: 3
Apr 30, 2021 at 12:35 review Close votes
May 12, 2021 at 3:02
Apr 30, 2021 at 12:16 comment added Ilya Bogdanov In this case, for $K=5$ you will get an embedding of $K_5$ in the plane.
Apr 30, 2021 at 10:59 comment added TheVal @JosephO'Rourke Yep, yours is a solution, as the partition share the central vertex $x$. I should add another question with a stricter requirement, that all $p_i$ must share at least one edge with all other parts.
Apr 30, 2021 at 10:48 history edited TheVal CC BY-SA 4.0
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Apr 30, 2021 at 10:47 comment added TheVal @JukkaKohonen Absolutely yes, and I'm trying to visualize Joseph solution, I'll tell in the comments asap. If that works, it surely will be a neat answer!
Apr 30, 2021 at 10:38 comment added Jukka Kohonen Thanks, it is clearer now. Is it also so that the INTERIORS of the parts must be pairwise disjoint? (Otherwise surely they cannot share a vertex or an edge.) Still it seems Joseph's comment would provide an easy solution, so is this what you are after, or something else?
Apr 30, 2021 at 10:35 comment added TheVal @JukkaKohonen I've edited my question. I apologize for the bad writing, and I hope I clarified a bit. I could add some diagrams if it helps to drive the point of my question home
Apr 30, 2021 at 10:33 history edited TheVal CC BY-SA 4.0
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Apr 30, 2021 at 9:52 comment added TheVal @JukkaKohonen Terribly sorry. I'll clarify right away: YES $p_i$s are pairwise disjoint, and YES $p_i$s are poligonal parts (subsets) of $P$, and $\mathcal{Z}_K(P)$ is a partition into $K$ parts and NOT the set of all such partitions
Apr 30, 2021 at 6:18 comment added Jukka Kohonen Fedor, my guess is the OP means that their interiors must have empty intersection. This is another thing that should be clarified in the question.
Apr 30, 2021 at 5:56 comment added Fedor Petrov Wait, for $K=2$ the intersection of parts must be empty but they should have a common edge or vertex. How is this possible?
Apr 30, 2021 at 5:40 history edited Francesco Polizzi
edited tags
Apr 30, 2021 at 5:26 history edited Francesco Polizzi CC BY-SA 4.0
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Apr 30, 2021 at 3:12 comment added Wlod AA Let $P=S^n$ be the boundary of an $(n+1)$-simplex. The intersection of the $n+2$ main faces is empty, the union is the entire $S^n$, and every two intersect at their common $(n-1)$-dimensional face.
Apr 30, 2021 at 3:05 history edited Wlod AA CC BY-SA 4.0
partitions are something else
Apr 30, 2021 at 0:37 comment added Joseph O'Rourke Why is this not a solution? Let $x$ be any strictly interior point of $P$. Draw spokes from $x$ to the boundary of $P$, with adjacent spokes separated by $2 \pi /K$. The resulting pieces $p_i$ each share $x$.
Apr 30, 2021 at 0:21 comment added Jukka Kohonen Could you clarify your notation slightly? I believe you want to specify that the $p_i$ are pairwise disjoint, instead of saying the intersection of all of them is empty. Also I get confused by what is a "part", what is "partition" and what is "partitioning". Are $p_i$ polygonal parts (subsets) of $P$? Is ${\cal Z}_K(P)$ a partition into $K$ such polygonal parts, as your notation says, or is it the set of all such partitions?
Apr 29, 2021 at 23:53 history asked TheVal CC BY-SA 4.0